SATHEE: Solutions and Colligative Properties

Solutions and Colligative Properties - Result Question 54

####25. An organic compound $(C_{x} H_{2 y} O_{y})$ was burnt with twice the amount of oxygen needed for complete combustion to $C O_{2}$ and $H_{2} O$ . The hot gases when cooled to $0^{\circ} C$ and $1 a t m$ pressure, measured $2.24 L$ . The water collected during cooling weight $0.9 g$ . The vapour pressure of pure water at $20^{\circ} C$ is $17.5 m m H g$ and is lowered by $0.104 m m$ when $50 g$ of the organic compound are dissolved in $1000 g$ of water. Give the molecular formula of the organic compound.

$(1983, 5 M)$

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Solution:

From lowering of vapour pressure information :

$\begin{aligned} \frac{0.104}{17.5} = χ_{2} & = \frac{n_{2}}{n_{1} + n_{2}} \Rightarrow & \frac{n_{1}}{n_{2}} + 1 & = 168.27 \Rightarrow & \frac{n_{1}}{n_{2}} & = 167.27 \Rightarrow & \frac{1000}{18} \times \frac{M}{50} & = 167.27 \Rightarrow & M & = 150 g / m o l \end{aligned}$

Also, the combustion reaction is :

$C_{x} H_{2 y} O_{y} + x O_{2} ⟶ x C O_{2} + y H_{2} O$

$∵ 18 y g$ of $H_{2} O$ is produced from 1.0 mole of compound.

$∴ 0.9 g$ of $H_{2} O$ will be produced from $\frac{0.9}{18 y} = \frac{1}{20 y}$ mol

$\Rightarrow$ At the end, moles of $O_{2}$ left $= \frac{x}{20 y}$

moles of $C O_{2}$ formed $= \frac{x}{20 y}$

$\Rightarrow$ Total moles of gases at STP $= \frac{2 x}{20 y} = \frac{2.24}{22.4}$

$\Rightarrow x = y$

$\Rightarrow$ Molar mass; $150 = 12 x + 2 x + 16 x = 30 x$

$\Rightarrow x = \frac{150}{30} = 5$

$\Rightarrow$ Formula $= C_{5} H_{10} O_{5}$

Solutions and Colligative Properties - Result Question 54

Solution:

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Solutions and Colligative Properties - Result Question 54

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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